Potential Difference between two charged metal plates A and B



Batteries and Potential Difference.

A battery is a device that stores positive charge on one of its terminals and negative charge on the other terminal.

Electric Potential Energy.

Potential energy can be defined as the capacity for doing work which arises from position.

In the electrical case, a charge will exert a force on any other charge and potential energy arises from any collection of charges.

For example, if a positive charge Q1 is fixed at some point in space, any other positive charge which is brought close to it will experience a repulsive force and will therefore have potential energy. The potential energy of a test charge Q2 in the vicinity of this source charge will be:

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Electric Potential.

At a point in space, the electric potential is the potential energy per unit of charge that is associated with a static electric field. It is a scalar quantity, typically measured in volts.

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Potential at a point is defined as the work done in moving a charge of 1 coulomb from infinity to that point. Symbol for potential at the point A is

VA

The difference in electrical potential between two points is known as the potential difference or voltage.

Potential Difference between two charged metal plates A and B

A B

+ -

The electric field that exists between two metal plates is uniform except at the edges. Check the diagrams for the electric field lines.

If an electron ( -ve charge ) were moved from the positive plate A to the negative plate B, work would have to be done to overcome the force of repulsion between the -ve electron and the -vely charge metal plate at B.

This work would be stored in the election as potential energy as long as the election remains at plate B. This is also true if a +ve charge is moved from plate B to A.

Because of this change in potential energy we say that there is a difference in potential or potential difference ( p.d. ) between the plates.

The potential difference in this context is an indication of the amount of work that would be done in moving a charge between two points.

The S.I. unit of potential difference is called the Volt and the potential difference is also called voltage.

The unit Volt is defined as follows : There is a potential difference of 1 Volt between two points when 1 Joule of work has to be done to move 1 Coulomb of charge between the two points.

VA B is symbol for the potential difference between A and B

Note also VAB = VA - VB

Relationship between potential difference V and work done (energy) is given by

WORK DONE = Q . V.

Relationship between Electric Field Intensity ( E ) and Potential V at a distance of r(m) from a point charge is given by

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Relationship between Electric Field Intensity ( E ) and Potential difference V between two parallel charge metal plates a distance d (m) apart is given by

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Also it is important to note that Potential or Potential difference is a scalar quantity and this is always the case the

if Q is -ve then the potential is negative and

if Q is +ve then the potential is positive.

Electric Dipole: Equal and opposite point charges a distance r (m) apart.

Q1 Q2

r (m)

Example 1.An electric dipole has a positive charge of 20μC and a length of 0.5mm.

Calculate (i) the electric force between the two charges

(ii) Electric field intensity at the midpoint (A) of the dipole.

(iii) The potential at the midpoint.

Q1 A Q2

r (m)

Q1 = + 20x10-6C Q2 = - 20x10-6 C

(i) Use Coulomb’s Law to calculate the value of the force i.e. magnitude

F = 1 (Q1( (Q2 (

4πε r2

= 9 x 109 20 x 10-6 x 20 x 10-6

(5 x 10-4) 2

= 1.44 x 107 N

(ii) To calculate the total electric field intensity at the midpoint of the dipole we will calculate the field strength due to each point charge individually first and then calculate the total by adding the two values together remembering that there are each vectors and so they have to be added as such.

E1A = The electric field intensity at the midpoint A of the dipole due to the charge Q1 a distance

2.5 mm from A

E2A = The electric field intensity at the midpoint A of the dipole due to the charge Q2 a distance

2.5 mm from A

As E is a vector quantity we need to calculate its magnitude and its direction

(E ( = k (Q(

r2

(E1A ( = k (Q1(

r2

= 9x109(20 x10-6)

(2.5 x10-4)2

= 2.88 x 1012 N C -1 Acting Away from Q1

(E2A ( = k (Q2(

r2

= 9x109(20 x10-6)

(2.5 x10-4) 2

= 2.88 x 1012 N C -1 Acting towards Q2

So the directions of E1A and E2A are the same i.e. away from Q1 and towards Q2 are the same direction so the total value of

EA = 2.88 x 1012 N C -1 + 2.88 x 1012 N C -1

EA = 5.66 x 1012 N C -1 Towards Q2

Recall adding two vectors in the same direction.

(iii) To calculate the potential at A

since Q1 is a positive charge it causes a positive potential

V1A = E1A . r = 2.88 x 1012 ( 2.5 x10-4)

= 7.2 x 108 Volts

As Q2 is a negative charge it causes a negative potential.

V2A = E2A . r = - 2.88 x 1012 (2.5 x10-4)

= - 7.2 x 108 Volts

VA = V1A + V2A = 0 V

Problem Sheet Electrostatics.

1. Calculate (a) The electric field intensity in air at a distance of 30 cm from a point charge of 5 nC. (b) The force acting on a charge of 400μC placed at 30cm from the original charge.

2. A dipole has a positive charge of 6μC and a length of 0.2mm

Calculate (a) The force acting on the -ve charge (b) The electric field intensity and the electric potential at the midpoint of the dipole. (c) The electric field intensity at a distance of 0.5mm beyond the +ve charge.

3. Two parallel plates have a p.d. of 500 kV placed across them. The plates are 20cm apart. Calculate (a) The electric field intensity between the plates (b) The force acting on an electron placed between the plates given that e = 1.6 x 10-19 C.

(c) How long will it take an electron starting from one plate to reach the other plate me ( mass of electron ) = 9 x 10-31 kg.

4. A dipole has a positive charge of 1.6 x 10-19C and a length of 2 x 10 -10 m

Calculate (a) The force acting on the -ve charge (b) The electric field intensity and the electric potential at the midpoint of the dipole ( point A ). (c) The electric field intensity and the electric potential at a distance of 10 x 10 -10 m beyond the +ve charge ( point B ).

5. An electric field is maintained between two parallel metal plates by a p.d. of 1000 V. A charge of 1 x 10 -15 C is moved between the plates by the forces of the field which exists between the plates. The plates are 20 mm apart.

(a) How much work is done in moving the charge from one plate to another

(b) Calculate the field strength in N C-1

(c) Calculate the field strength in V m-1

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